# Math Help - Limit

1. ## Limit

Find $lim_{x\to\infty} x[(1+\frac{1}{x})^x-e]$ [Suggestion: Let $u=\frac{1}{x}$, and take the limit as $u\to0$.]

2. Setting $x=\frac{1}{u}$ and applying several times l'Hopital's rule is ...

$\displaystyle \lim_{u \rightarrow 0} \frac{(1+u)^{\frac{1}{u}} - e}{u}= \lim_{n \rightarrow 0} \frac{(1+u)^{\frac{1}{u}}}{1+u}\ \frac {u- \ln (1+u)}{u^{2}} = e\ \lim_{u \rightarrow 0} \frac{1- \frac{1}{1+u}}{2u} =$

$\displaystyle = e \lim_{u \rightarrow 0} \frac{1}{2\ (1+u)^{2}} = \frac{e}{2}$

Kind regards

$\chi$ $\sigma$